Question

How do I use synthetic division to find the other zeros

Answer 1

Given the function below

$f(x)=x^4-11x^3+40x^2-48x^{}$

Where

$x=3\text{ is a zero of the function.}$

To find the other zeros,

$\begin{gathered} x=3x \\ x-3=0\text{ is a factor} \end{gathered}$

And x is common, factor out x, i.e

$\begin{gathered} f(x)=x^4-11x^3+40x^2-48x^{} \\ f(x)=x(x^3-11x^2+40x-48) \end{gathered}$

Divde the function by the factor x - 3

The quotient of the function after dividing by x - 3 is

$f(x)=(x-3)(x^3-8x^2+16x)$

Factor out x

$\begin{gathered} f(x)=(x-3)(x^3-8x^2+16x) \\ f(x)=x(x-3)(x^2-8x+16) \end{gathered}$

Factorize the remaining equation

$\begin{gathered} f(x)=x(x-3)(x^2-8x+16) \\ f(x)=x(x-3)(x^2-4x-4x+16) \\ f(x)=x(x-3)\mleft\lbrace x(x-4\mright)-4(x-4)\} \\ f(x)=x(x-3)\mleft\lbrace(x-4\mright)(x-4)\} \\ f(x)=x(x-3)(x-4)^2 \end{gathered}$

To find the zeros of the above factored function, using the zero factor principle

$\begin{gathered} xy=0 \\ x=0\text{ and y}=0 \end{gathered}$

$\begin{gathered} x(x-3)(x-4)^2=0 \\ x=0 \\ x-3=0 \\ x=3 \\ x-4=0 \\ x-4=0 \\ x=4 \\ x=0,3,4(\text{twice)} \end{gathered}$

Hence, the othe two zeros of the function f(x) are

$x=0,x=4$